Let be a differentiable function that satisfies the relation . If , then is equal to
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Answer: 03
Solution
Setting y = 0: f(x) = f(x) + f(0) - 1, so f(0) = 1.
The functional equation f(x+y) = f(x) + f(y) - 1 can be rewritten by letting g(x) = f(x) - 1, giving g(x+y) = g(x) + g(y), a Cauchy equation. Since f is differentiable, g(x) = cx, so f(x) = cx + 1.
f'(0) = c = 2, so f(x) = 2x + 1.
f(-2) = 2(-2) + 1 = -3, so |f(-2)| = 3.
The functional equation f(x+y) = f(x) + f(y) - 1 can be rewritten by letting g(x) = f(x) - 1, giving g(x+y) = g(x) + g(y), a Cauchy equation. Since f is differentiable, g(x) = cx, so f(x) = cx + 1.
f'(0) = c = 2, so f(x) = 2x + 1.
f(-2) = 2(-2) + 1 = -3, so |f(-2)| = 3.
[JEE Main 2023 · 29 Jan · Shift 1]